Maximum Product of Two Elements in an Array

Maximum Product of Two Elements in an Array - Problem

Given an array of integers nums, you will choose two different indices i and j of that array.

Return the maximum value of (nums[i]-1) * (nums[j]-1).

Input & Output

Example 1 — Basic Case

$ Input: nums = [3,4,5,2]

› Output: 12

💡 Note: Choose indices 1 and 2: (4-1) * (5-1) = 3 * 4 = 12

Example 2 — Minimum Size

$ Input: nums = [1,5]

› Output: 0

💡 Note: Only two elements: (1-1) * (5-1) = 0 * 4 = 0

Example 3 — All Same Values

$ Input: nums = [3,3,3,3]

› Output: 4

💡 Note: All elements equal: (3-1) * (3-1) = 2 * 2 = 4

Constraints

2 ≤ nums.length ≤ 500
1 ≤ nums[i] ≤ 10³

Visualization

Tap to expand

Asked in

a Amazon 15 M Microsoft 8

The key insight is that maximum product comes from the two largest elements in the array. Best approach is the one-pass solution tracking the largest and second-largest elements. Time: O(n), Space: O(1)

Common Approaches

✓ Brute Force - Check All Pairs

⏱️ Time: O(n²) Space: O(1)

Iterate through all pairs of different indices (i,j) and calculate (nums[i]-1)*(nums[j]-1) for each pair. Keep track of the maximum product found.

One Pass - Track Two Largest

⏱️ Time: O(n) Space: O(1)

Iterate through the array once while keeping track of the largest and second largest elements. This avoids the overhead of sorting while achieving optimal result.

Sort and Pick Top Two

⏱️ Time: O(n log n) Space: O(1)

Sort the array in descending order, then take the first two elements. Calculate (first-1) * (second-1) to get the maximum product directly.

Brute Force - Check All Pairs — Algorithm Steps

Nested loops through all pairs (i,j) where i ≠ j
Calculate product (nums[i]-1) * (nums[j]-1)
Track maximum product seen so far

Visualization

Tap to expand

Step-by-Step Walkthrough

Check Pairs

Try all combinations of two different indices

Calculate

Compute (nums[i]-1) * (nums[j]-1) for each pair

Track Maximum

Keep the highest product found

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int solution(int* nums, int numsSize) {
    int maxProduct = 0;
    
    for (int i = 0; i < numsSize; i++) {
        for (int j = i + 1; j < numsSize; j++) {
            int product = (nums[i] - 1) * (nums[j] - 1);
            if (product > maxProduct) {
                maxProduct = product;
            }
        }
    }
    
    return maxProduct;
}

int main() {
    char line[1000];
    fgets(line, sizeof(line), stdin);
    
    // Parse JSON array manually
    int nums[1000];
    int count = 0;
    char* token = strtok(line + 1, ",]"); // Skip opening bracket
    
    while (token != NULL) {
        nums[count++] = atoi(token);
        token = strtok(NULL, ",]");
    }
    
    int result = solution(nums, count);
    printf("%d\n", result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

Nested loops check all n*(n-1) pairs

⚡ Linearithmic

Space Complexity

O(1)

Only using variables for tracking maximum

✓ Linear Space

25.0K Views

Medium Frequency

~10 min Avg. Time

850 Likes

Ln 1, Col 1

Smart Actions

💡 Explanation

AI Ready

💡 Suggestion Tab to accept Esc to dismiss

// Output will appear here after running code

Code Editor Closed

Click the red button to reopen

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force - Check All Pairs — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Select Compiler